A) 200g/L
B) 19.2g/L
C) 17.54g/L
D) 16.7g/L
Correct Answer: C
Solution :
\[{{V}_{1}}=1\,{{L}_{2}}{{V}_{3}}=100\,mL=0.1L\] \[\frac{{{W}_{1}}}{{{m}_{1}}{{V}_{1}}}=\frac{{{W}_{2}}}{{{m}_{2}}{{V}_{2}}}\] \[\frac{{{W}_{1}}}{60\times 1}=\frac{10}{342\times 0.1}\] \[{{W}_{1}}=17.54\,g/L\]You need to login to perform this action.
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