Question

What is the weight (in grams) of \[N{a_2}C{O_3}\] molar mass is \[106\] present in \[250ml\], of its \[0.2M\] solutions?
A.\[0.53\]
B.\[5.3\]
C.\[1.06\]
D.\[10.6\]

Answer 1

Hint: Molarity can be calculated from the number of moles of solute and volume of solution in liters. The number of moles of solute can be calculated by dividing the weight of substance by molar mass. Molarity was already given and the weight of sodium carbonate can be calculated from the formula of molarity.

Formula used:
\[w = \dfrac{{M \times m \times v}}{{1000}}\]
w is weight of sodium carbonate in grams
M is molarity
M is molar mass of sodium carbonate
v is volume of solution in ml

Complete answer:
Molarity is one of the units used to express the concentration of a substance. It is also known as molar concentration. It was represented by M and has the units of \[mol{\left( {lit} \right)^{ - 1}}\]. It can be calculated from the volume of solution and number of moles of solute.
Given that sodium carbonate has a molar mass of \[106gm{\left( {mol} \right)^{ - 1}}\] and the molarity is \[0.2M\].
The volume of solution is \[250ml\]
Substitute the values in the above formula
\[w = \dfrac{{106 \times 0.2 \times 250}}{{1000}}\]
Upon simplifying the values, we will get
\[w = 5.3\]
Thus, the weight of sodium carbonate is \[5.3grams\] whose molar mass is \[106\] present in \[250ml\], of its \[0.2M\] solution.
Option B is the correct one.

Note:
While calculating the molarity or weight of substance present in grams by using the formula of molarity, the volume of solution must be in litres. If it is present in milliliters then it should be multiplied with \[1000\] as a litre is equal to \[1000ml\].