Search results
5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water. Q. A 5 % solution (by mass) of cane sugar in water has freezing point of 271 K and freezing point of pure water is 273.15 K. The freezing point of a 5 % solution (by mass) of glucose in water is : Q.
Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W 2) = 5 g, and mass of solvent (W 1) = 95 g. 5 % glucose solution means that mass of glucose = W W 2 ′ = 5g, and mass of solvent = W W 1 ′ = 95 g. Molar mass of glucose (C 6 H 12 O 6) = M (M 2 ′) = 180 g mol -1.
13 gru 2017 · Molar mass of sugar (C 12 H 22 O 11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol− 1.
A 5 % solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5 % glucose in water if freezing point of pure water is 273.15 K. Molar masses of glucose and sucrose are 180 g/mol and 342 g/mol respectively. 100 g of solution will contain 5 g of glucose or 5 g of sucrose. Was this answer helpful?
Now, we will calculate the molality of glucose. A 5% solution of cane sugar means 5g glucose is dissolved in 95g of water. Therefore, Tglu cos e = 269.04K T g l u cos e = 269.04 K, Which is close to option (B). So, the correct answer is “Option B”. Note: Convert the temperatures into kelvin. Take weight of solution in Kilograms.
2 lis 2018 · A 5 percent solution (by mass) of cane-sugar (M.W. 342) is isotonic wirh 0.877% solution of substance X. Find the molecular weight of X. Play Quiz Games with your School Friends. Click Here.
As, 5% of solute is present in the solution thus, 5 gm of solute is present in 95 gm of solvent (for both sucrose and glucose). The molar masses of glucose and sucrose are 180 gm/mol and 342 gm/mol respectively.
