# How much KI is present in .836 L of a 1.82 M solution, assuming the solvent is H_2O?

Mar 11, 2016

The solution contains 252 g of $\text{KI}$.

#### Explanation:

Step 1. Calculate the moles of $\text{KI}$.

Your solution contains 1.82 mol of $\text{KI}$ in 1 L of solution.

So, in 0.836 L of solution you have

0.836 color(red)(cancel(color(black)("L solution"))) × "1.82 mol KI"/(1 color(red)(cancel(color(black)("L solution")))) = "1.522 mol KI"

Step 2. Convert moles of $\text{KI}$ to grams of $\text{KI}$

$\text{1 mol KI}$ has a mass of $\text{(39.10 + 126.90) g = 166.00 g}$.

Hence,

$\text{Mass of KI" = 1.522 color(red)(cancel(color(black)("mol KI"))) × "166.00 g KI"/(1 color(red)(cancel(color(black)("mol KI")))) = "252 g KI}$