Question

How many milliliters of $0.45\text{MBaC}{\text{l}}_{\text{2}}$ solution contain 15.0 g of $BaC{l}_2(208gmo{l}^{-1})$

• A. 110.7mL
• B. 124.6mL
• C. 135.8 mL
• D. 160.3 Ml

Answer 1

Answer: (D)

Let volume of the solution containing 15,0 g $BaC{l}_3$ be VmL. $M_1{V}_1=M_2{V}_2$ $V\times 0.45=\frac{15}{208}\times 1000$ $V=160.256mL\approx 160.3mL$