Question #8af7e

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Question #8af7e

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Answer 1 · 2015-12-11T15:40:47

$m_{N a C l} = 394.2 g$ $m_(NaCl)=394.2g$

Explanation:

The concentration is related to the mass through the number of mole:

$C_{M} = \frac{n}{V}$ $C_M=n/V$

and the number of mole is related to the mass by:

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

Therefore, we can first calculate the number of mole from the concentration and the volume by:

$n = C_{M} \times V = 4.50 \frac{m o l}{L} \times 1.50 L = 6.75 m o l$ $n=C_MxxV=4.50(mol)/(cancel(L))xx1.50cancel(L)=6.75mol$

Now from the number of mole we can find the mass by:

$m = n \times M M = 6.75 m o l \times 58.4 \frac{g}{m o l} = 394.2 g$ $m=nxxMM=6.75cancel(mol)xx58.4 g/(cancel(mol))=394.2g$

Thus we will need to weigh $394.2 g$ $394.2g$ of $N a C l$ $NaCl$ and add it to $1.50 L$ $1.50L$ of water to prepare $4.50 M$ $4.50M$ solution of sodium chloride.

Here is a video that will show you how to prepare a solution in the lab:
Lab Demonstration | Solution Preparation & Dilution.

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Question #8af7e

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