What is the molarity if you dissolve 15 g of $F e C l_{3}$ $FeCl_3$ to make up 250 mL solution?

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Answer 1 · 2016-03-07T21:48:03

0.37 M

First calculate the molar mass of $F e C l_{3}$ $FeCl_3$ :

$55.85 + 3 \cdot 35.45 = 162.20 g / m o l$ $55.85+3*35.45=162.20 g//mol$

Then calculate how many moles are 15 g of $F e C l_{3}$ $FeCl_3$ :

$n = \frac{m}{M}$ $color(red)(n=m/M$

$n = \frac{15}{162.20} = 0.0925 m o l$ $n=15/162.20=0.0925 mol$

Then calculate the concentration:

$c = \frac{n}{v}$ $color(red)(c=n/v$

$n = 0.0925 m o l / 0.250 L = 0.37 m o l / L (or 0.37 M)$ $n=0.0925mol//0.250L=0.37 mol//L (or 0.37 M)$

The concentration is given with two representative digits which are the same ones of 15 grams.

What is the molarity if you dissolve 15 g of FeCl3FeCl_3 to make up 250 mL solution?