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  4. How many milliliters of 0.45 textMBaC textl text2 solution contain 15.0 g of BaCl2(208 g mol-1)

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

VMMC MedicalVMMC Medical 2013

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 $