Question

A solution contains a compound with a molar mass of 26.36 g/mol. How many gramsof this compound are contained if the solution has a volume of 1,258.1 mL and amolarity of 1.1 M?

Answer 1

36.48 grams

Step-by-step explanation:

Molarity = 1.1 M

Volume = 1258.1 mL = 1.2581 L

Molar mass = 26.36 g/mol

Firstly you need to know the number of moles of the compound.

The number of mole of the compound is calculated as follows:

`\begin{gathered} \text{Molarity }=(Moles)/(Volume) \\ 1.1\text{ M }=\frac{Moles}{1.2581\text{ L}} \\ \text{Moles }=1.1\text{ M }*1.2581\text{ L} \\ \text{Moles }=1.38391\text{ mol} \end{gathered}`

The grams of this compound would be:

`\begin{gathered} \text{moles }=\frac{Mass}{Molar\text{ Mass}} \\ 1.38391\text{ mol }=\frac{Mass}{26.36\text{ g/mol}} \\ \text{Mass }=1.38391\text{ mol }*26.36\text{ g/mol} \\ \text{Mass }=36.4798676\text{ grams} \\ \text{Mass }\approx36.48\text{ grams} \end{gathered}`