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The boiling point of water (100^@ C) becomes, 100.52^@ C if 3 g of non-volatile solute is dissolved is 200 ml of it: Calculate the molecular weight of the solid. (K_b = 0.6 K/m) |
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Answer» Solution :`DELTA T_b = 100.52 -100=0.52 ^@ C K` ` W_B =3g` `W_A =200 ml` ` M_B=?` ` K_b = 0.6 K//m` ` DeltaT_b =K_b xx(W_b xx 1000 )/( M_B W_A)` ` 0.52 =( 0.6 xx 3xx 1000)/( 200xx 0.52 )` `=( 90 )/( 0.52)` `=173.07//Mol .` |
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