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Question

Chemistry Question on Solutions

What is the relation between molar mass of solute and boiling point elevation of solution?

A

M2 = 1000 △TbW2KbW1\frac {1000\ △T_bW_2}{K_bW_1}

B

M2 = 1000 KbW2TbW1\frac {1000\ K_bW_2}{△T_bW_1}

C

M2 = TbW11000 KbW2\frac {△T_bW_1}{1000\ K_bW_2}

D

M2 = 1000 KbW1TbW2\frac {1000\ K_bW_1}{△T_bW_2}

Answer

M2 = 1000 KbW2TbW1\frac {1000\ K_bW_2}{△T_bW_1}

Explanation

Solution

The relation between the molar mass of the solute (M2) and the boiling point elevation of the solution (△Tb) can be given by the equation:
△Tb = Kb.M2.W2W1\frac {K_b . M_2 . W_2}{W_1}
where:
△Tb is the boiling point elevation of the solution,
Kb is the molal boiling point elevation constant,
M2 is the molar mass of the solute,
W2 is the weight (mass) of the solute,
and W1 is the weight (mass) of the solvent.
To convert the equation into a more convenient form, we can multiply both sides by 1000:
1000 . △Tb = 1000.Kb.M2.W2W1\frac {1000 .K_b . M_2 .W_2}{W_1}
Simplifying, we have:
M2 = 1000.Kb.W2Tb.W1\frac {1000 . K_b . W_2}{△T_b . W_1}
We can see that the correct relation is option (B): M2 = 1000.Kb.W2Tb.W1\frac {1000 . K_b . W_2}{△T_b . W_1}