A) 110.7mL
B) 124.6mL
C) 135.8 mL
D) 160.3 Ml
Correct Answer: D
Solution :
Let volume of the solution containing 15,0 g \[BaC{{l}_{3}}\]be VmL. \[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\] \[V\times 0.45=\frac{15}{208}\times 1000\] \[V=160.256\,mL\approx 160.3\,mL\]
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