What is the molality of a solution that contains 75.2 grams of AgClO4 in 885 grams of benzene? Question 5 options: 0.41 m 4.10 m 8.20 m 0.83 m

I took the test and got this question right! Here is the work.

molality = moles solute / kg solvent

moles AgClO4 = 75.2 g / 207.389 g/mol

= 0.3626 mol

molality = 0.6326 mol / 0.885 kg

= 0.410 m

The asnwer is 0.41.