Question

How many grams of solute are present in $795$ mL of $0.870$ M KBr?

Answer 1

Hint: We can solve this problem with the concept of molarity. Molarity of a solution is the number of moles of the solute present in one litre of the solution. Molarity of a solution is a measure to express its concentration.

Complete step by step answer:
We need to find the weight of solute in grams present in $795$ mL of $0.870$ M solution of KBr. The equation for molarity can be written as,
$M = \dfrac{{{w_B} \times 1000}}{{{M_B} \times V}}$
Where M is the molarity of the solution, ${w_B}$ is the weight of solute in grams, ${M_B}$ is the molar mass of solute in $g/mol$ and V is the volume of solution in mL.
Given that, Molarity of the solution, $M = 0.870M$
Volume of solution, $V = 795mL$
We need to find ${w_B}$ .
The solute is potassium bromide (KBr). Its molar mass is $119.002g/mol$ .
Hence ${M_B} = 119.002g/mol$
Let us rewrite the above equation as,
${w_B} = \dfrac{{M \times {M_B} \times V}}{{1000}}$
Now let us substitute the values to calculate ${w_B}$ .
${w_B} = \dfrac{{0.870 \times 119.002 \times 795}}{{1000}} = 82.31g$
Hence the weight in grams of solute present in $795$ mL of $0.870$ M KBr is $82.31g$ .

Additional information-
The term $\dfrac{{{w_B}}}{{{M_B}}}$ gives the number of moles of the solute. We have seen that the molar mass of KBr is $119.002g/mol$ . This means that the weight of one mole of KBr is $119.002$ grams.

Note:
We can also write the equation of molarity as,
$M = \dfrac{{{w_B}}}{{{M_B} \times V}}$
But when we are using this equation, we should not forget to convert the volume from mL to litre. Usually in the problems, volume will be given in mL. Hence the equation which we used to solve the problem is more convenient.