Question

How many grams of sodium sulphate $\left( N{{a}_{2}}S{{O}_{4}} \right)$ are required to make 4.5 litres of a 3.0 M solution?

Answer 1

Hint To solve this illustration we need to have a basic idea about the molarity of a solution. Knowing the molarity can help us solve this question properly. Also, molar mass of Na is 23 g/mol, molar mass of S is 32 g/mol and molar mass of O is 16 g/mol.

Complete step by step solution:
Let us see what we mean by molarity and solve the illustration in accordance with it;
Molarity- Molarity is the number of moles of solute per litre of solution. It is also known as molar concentration and is represented as M (mol/L). the equation can be written as;
$M=\dfrac{n}{V}$
M = molarity
n = number of moles of solute
V = volume of solution
Number of moles-
The number of moles of any substance is the ratio of the given mass of the substance to the molar mass of the substance. The equation is represented as,
$n=\dfrac{{{m}_{g}}}{{{m}_{m}}}$
where,
${{m}_{g}}$ = given mass of substance
${{m}_{m}}$ = molar mass of substance
The illustration can be solved as;
Given data-
Volume = 4.5 L
Molarity = 3.0 M
Thus, using above equations,
$\begin{aligned} & M=\dfrac{n}{V} \\\ & n=VM \\\ & n=4.5L\times 3.0M \\\ & n=13.5mol \\\ \end{aligned}$
Now,

& n=\dfrac{{{m}_{g}}}{{{m}_{m}}} \\\ & {{m}_{g}}=n\times {{m}_{m}} \\\ & {{m}_{g}}=13.5mol\times \left[ \left( 2\times 23 \right)+\left( 1\times 32 \right)+\left( 4\times 16 \right) \right]g/mol \\\ & {{m}_{g}}=1917g \\\ \end{aligned}$$ **Note:** Do note to use proper units while solving the numerical illustration. And the molar mass is calculated as the summation of the product of the atomic masses and the subscripts involved while formulation; we have directly calculated that while solving the above problem.