A) \[200g/lit\]
B) \[19.2g/lit\]
C) \[17.54g/lit\]
D) \[16.7g/lit\]
Correct Answer: C
Solution :
\[{{V}_{1}}=1litre\] and \[{{V}_{2}}=100mlit\,=0.1\,litre\] \[\frac{{{\omega }_{1}}}{{{m}_{1}}{{V}_{1}}}=\frac{{{\omega }_{2}}}{{{m}_{2}}{{V}_{2}}}\] or \[\frac{{{\omega }_{1}}}{60\times 1}=\frac{10}{342\times 0.1}\] or \[{{\omega }_{1}}=17.54g/L\]
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