
Molarity and molality of a solution of a liquid (mol wt = 50) in aqueous solution is 9 and 10 respectively. What is the density of solution?
(A)- 1 g/cc
(B)- 0.95 g/cc
(C)- 1.05 g/cc
(D)- 1.35 g/cc
Answer
448.6k+ views
Hint: Density of a solution is a measure of its mass present per unit volume. It is given as
\[\text{Density = }\frac{\text{mass of solution}}{\text{volume of solution}}\]
Density is generally expressed in g/cc (or g /ml). Its S.I unit is kg/${{m}^{3}}$.
Molality of a solution gives the mass of the solute present in one kg of the solvent. It is calculated as
\[\text{Molality (m) = }\frac{\text{no}\text{. of moles of solute}}{\text{weight of solvent (in kg)}}=\frac{\text{mass of solute (in g)}}{\text{molar mass of solute}}\times \frac{1000}{\text{weight of solvent (in g)}}\]
Molarity of a solution is the mass of the solute present in per liter of the solution. It can be calculated using the following relation
\[\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution (in L)}}=\frac{\text{mass of solute (in grams)}}{\text{molar mass of solute}}\times \frac{1000}{\text{volume of solution (in mL)}}\]
Complete step by step solution:
We have been given the molar mass of the liquid dissolved in the aqueous solution = 50 g
To find the density of the solution we require the total mass and total volume of the solution, i.e.
Total mass of the solution, m = mass of the solute + mass of the solvent
Total volume of the solution, V = volume of the solute + volume of the solvent
We know that molality of a solution is the number of moles of solute dissolved in per kg of the solvent.
Given molality of the aqueous solution = 10 mol $k{{g}^{-1}}$.
10 molal solution means that 10 moles of the given liquid are dissolved in 1000 g (1 kg) of the solvent. Then, 1 mole of the liquid solute will be present in $\frac{1000}{10}$= 100 g of the solvent.
Thus, we have the mass of solvent = 100 g.
Now we already know that the mass of one mole of the liquid solute = 50 g.
Hence, the total mass of the solution (m) will be = 50 g + 100 g = 150 g.
Since molarity is the number of moles of a solute dissolved in one liter of solution. Given molarity of the aqueous solution = 9 mol $li{{t}^{-1}}$
9 molar solution contains 9 moles of the solute in 1000 ml (1 liter) of the solution, then 1 mole of the solute will be present in $\frac{1000}{9}$= 110 ml of the solution.
Therefore, we obtain the total volume of the solution (V) = 110 ml.
Let us now find the density of the solution.
\[\text{Density = }\frac{\text{mass of solution (m)}}{\text{volume of solution (V)}}\]
Substituting the value of total mass of the solution, m = 150 g and volume of the solution, V = 110 ml in the above equation, we get
\[\text{Density = }\frac{\text{150 g}}{\text{110 ml}}\text{=1}\text{.35}\]g/ml = 1.35 g/cc.
Hence, the correct option is (D).
Note: Note that 1 millimeter = 1 cubic centimeter. Do not get confused between molality and molarity. Molality does not involve the volume but mass of the solvent. Volume of the solution is obtained from the molarity.
\[\text{Density = }\frac{\text{mass of solution}}{\text{volume of solution}}\]
Density is generally expressed in g/cc (or g /ml). Its S.I unit is kg/${{m}^{3}}$.
Molality of a solution gives the mass of the solute present in one kg of the solvent. It is calculated as
\[\text{Molality (m) = }\frac{\text{no}\text{. of moles of solute}}{\text{weight of solvent (in kg)}}=\frac{\text{mass of solute (in g)}}{\text{molar mass of solute}}\times \frac{1000}{\text{weight of solvent (in g)}}\]
Molarity of a solution is the mass of the solute present in per liter of the solution. It can be calculated using the following relation
\[\text{Molarity}=\frac{\text{number of moles of solute}}{\text{volume of solution (in L)}}=\frac{\text{mass of solute (in grams)}}{\text{molar mass of solute}}\times \frac{1000}{\text{volume of solution (in mL)}}\]
Complete step by step solution:
We have been given the molar mass of the liquid dissolved in the aqueous solution = 50 g
To find the density of the solution we require the total mass and total volume of the solution, i.e.
Total mass of the solution, m = mass of the solute + mass of the solvent
Total volume of the solution, V = volume of the solute + volume of the solvent
We know that molality of a solution is the number of moles of solute dissolved in per kg of the solvent.
Given molality of the aqueous solution = 10 mol $k{{g}^{-1}}$.
10 molal solution means that 10 moles of the given liquid are dissolved in 1000 g (1 kg) of the solvent. Then, 1 mole of the liquid solute will be present in $\frac{1000}{10}$= 100 g of the solvent.
Thus, we have the mass of solvent = 100 g.
Now we already know that the mass of one mole of the liquid solute = 50 g.
Hence, the total mass of the solution (m) will be = 50 g + 100 g = 150 g.
Since molarity is the number of moles of a solute dissolved in one liter of solution. Given molarity of the aqueous solution = 9 mol $li{{t}^{-1}}$
9 molar solution contains 9 moles of the solute in 1000 ml (1 liter) of the solution, then 1 mole of the solute will be present in $\frac{1000}{9}$= 110 ml of the solution.
Therefore, we obtain the total volume of the solution (V) = 110 ml.
Let us now find the density of the solution.
\[\text{Density = }\frac{\text{mass of solution (m)}}{\text{volume of solution (V)}}\]
Substituting the value of total mass of the solution, m = 150 g and volume of the solution, V = 110 ml in the above equation, we get
\[\text{Density = }\frac{\text{150 g}}{\text{110 ml}}\text{=1}\text{.35}\]g/ml = 1.35 g/cc.
Hence, the correct option is (D).
Note: Note that 1 millimeter = 1 cubic centimeter. Do not get confused between molality and molarity. Molality does not involve the volume but mass of the solvent. Volume of the solution is obtained from the molarity.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE

The flightless birds Rhea Kiwi and Emu respectively class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE
