Question

The molality of a solution with x% mass by vol of $H_{2}S{O}_4$ solution is equal to 9 m. The weight of the solvent present in 1 L of this solution is 900 gm. What is the value of x?

Answer 1

Molality of the solution = 9 mol per kg
In 1 L solution 900 gm of solvent is present, moles of $H_{2}S{O}_4=9\times 0.900mol$
So, mass of $H_{2}S{O}_4=9\times 0.9\times 98=793.8gm$
So, in 100 ml solution mass of $H_{2}S{O}_4=79.38gm$