# Question #0e523

Dec 2, 2015

$9.56 \text{g}$

#### Explanation:

You have quoted molar heat capacities from your data book but you have used $m$ for mass in your expression for $Q$.

So we must work in moles and then convert back to grams using the idea that no. moles = mass/mass of 1 mole

Or in symbols:

$n = \frac{m}{M} _ r$

So:

$Q = n \times {C}_{v} \times \Delta T$

Note that $n$ is the no. moles.

Applying this first to helium:

The ${M}_{r}$ of helium is 4. (Same as the ${A}_{r}$).

So from our expression for $n$ we can write:

$Q = \frac{2}{4} \times 12.5 \times \Delta T \text{ } \textcolor{red}{\left(1\right)}$

Now for oxygen:

The ${M}_{r}$ of ${O}_{2}$ is 32 so:

$Q = \frac{m}{32} \times 20.9 \times \Delta T \text{ } \textcolor{red}{\left(2\right)}$

Now we can put $\textcolor{red}{\left(1\right)}$ equal to $\textcolor{red}{\left(2\right)} \Rightarrow$

$\frac{2}{4} \times 12.5 \times \cancel{\Delta T} = \frac{m}{32} \times 20.9 \times \cancel{\Delta T}$

$\therefore m = \frac{0.5 \times 12.5 \times 32}{20.9} = 9.56 \text{g}$

(I have used approximate ${M}_{r}$ values. You should use the ones you are given)