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Mole fraction of urea in 900 gram water is 0.05 density of solution is `1.2g//cm^(3)`. Find the molarity of solution. |
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Answer» Correct Answer - 2.98 or 2.99 `wt_(H_(2)O)=900gimpliesn_(H_(2)O)=(900)/(18)=50` Let moles of urea =y `impliesx_(urea)=(y)/(y+50)=0.05` `impliesy=(50)/(19)` `thereforeWt_(urea)=((50)/(19))xx60=(3000)/(19)` `thereforeWt_(solution)=(3000)/(19)+900=((201))/(19)xx100` `rho_("solution")=1.2g//cm^(3)` `impliesV_("solution")=(wt)/(rho)=((201)/(19))xx(100)/(1.2)` `impliesM_(urea)=(n_(urea))/(V_(ml))xx1000=(((50)/(19))xx1000)/(((201)/(19))xx(100)/(1.2))=((500)(1.2))/(201)=(600)/(201)cong2.985` Answer after rounding off =2.98 |
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