Question #c028e

1 Answer
Jan 27, 2017

The molality of the ${\text{CaCl}}_{2}$ is 1.14 mol/kg.

Explanation:

$\text{Molar mass of CaCl"_2·"2H"_2"O" = "110.98 g + 36.03 g" = "147.01 g}$

${\text{Mass of CaCl"_2 = 7.02 color(red)(cancel(color(black)("g CaCl"_2"·2H"_2"O"))) × "110.98 g CaCl"_2/(147.01 color(red)(cancel(color(black)("g CaCl"_2·"2H"_2"O")))) = "5.300 g CaCl}}_{2}$

$\text{Mass of H"_2"O" = 7.02 color(red)(cancel(color(black)("g CaCl"_2·"2H"_2"O"))) × ("36.03 g H"_2"O")/(147.01 color(red)(cancel(color(black)("g CaCl"_2·"2H"_2"O")))) = "1.720 g H"_2"O}$

$\text{Total mass of water" = "40.0 g + 1.720 g" = "41.72 g}$.

So, we have a solution of 5.300 g of ${\text{CaCl}}_{2}$ in 41.72 g of water.

${\text{Moles of CaCl"_2 = 5.300 color(red)(cancel(color(black)("g CaCl"_2))) × ("1 mol CaCl"_2)/(110.98 color(red)(cancel(color(black)("g CaCl"_2)))) = "0.047 75 mol CaCl}}_{2}$

$\text{Molality of CaCl"_2 = "moles of solute"/"kilograms of solvent" = "0.047 75 mol"/"0.041 72 kg" = "1.14 mol/kg}$