InterviewSolution
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InterviewSolution
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29.2% (W/W) HCI stock solution has a density of 1.25 g `mL^(-1)`. The molecular mass of HCI is 36.5 g `mol^(-1)` Calculate the volume (mL) of stock solution required to prepare 200 mL of 0.4 M HCIA. 2 mLB. 4 mLC. 8 mLD. 6 mL |
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Answer» Correct Answer - C Stock solution of HCI = 29.2 % (w/w) Thus, 29.2 g HCI are present in 100 g of the solution. As density of solution = 1.25 g `mL^(-1)`, Volume of 100 g of the solution = 100/1.25 mL Molar mass of HCI = 36.5 g `mol^(-1)`, `"Molarity of the solution" = (W_(2) xx 1000)/(M_(2) xx V (cm^(3)))` `29.2/36.5 xx 1.25/100 xx 1000 = 10 M` `{:(M_(1) V_(1),=,M_(2) V_(2)),(("stock solution"), ,("solution required")),(10 xx V_(1),=,0.4 xx 200):}` ` or V_(1) = 8 mL` |
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