Question

Find the molality of a solution made by mixing 85.3 g of 1.35 M HCl, 0.867 g of HCl and 89.1 g of
water.

Answer 1

Molar mass of HCl = 36.458 g/mol

Moles of HCl$=\frac{85.3}{36.458}=2.34 moles$

Volume of $HCl=\frac{2.34}{1.35}=1.73m^3$

Moles of water$=\frac{89.1}{18.02}=4.94 moles$

Molarity =moles/Volume

Molarity $=\frac{4.94}{1.73}=2.86M$