USING BOILING-POINT ELEVATION TO CALCULATE THE MOLALITY OF A SOLUTION What is the molality of an aqueous glucose solution if the boiling point of the solution at 1 atm pressure is 101.27 °C? The molal boiling-point-elevation constant for water is given in Table 11.4. TABLE 11.4 Molal Boiling-Point-Elevation Constants ([latex]K_{b}[/latex]) and Molal Freezing-Point-Depression Constants ([latex]K_{f}[/latex]) for Some Common Substances Substance [latex]K_{b}[/latex] [(°C · kg)/mol] [latex]K_{f}[/latex] [(°C · kg)/mol] Benzene ([latex]C_{6}H_{6}[/latex]) 2.64 5.07 Camphor ([latex]C_{10}H_{16}O[/latex]) 5.95 37.8 Chloroform ([latex]CHCl_{3}[/latex]) 3.63 4.70 Diethyl ether ([latex]C_{4}H_{10}O[/latex]) 2.02 1.79 Ethyl alcohol ([latex]C_{2}H_{6}O[/latex]) 1.22 1.99 Water ([latex]H_{2}O[/latex]) 0.51 1.86 STRATEGY Rearrange the equation for molal boiling-point elevation to solve for m: [latex]ΔT_{b} = K_{b} · m so m = \frac{ΔT_{b}}{K_{b}}[/latex] where [latex]K_{b}[/latex] = 0.51 (°C · kg)/mol and [latex]ΔT_{b}[/latex] = 101.27 °C - 100.00 °C = 1.27 °C.

m = [latex]\frac{1.27 °C}{0.51\frac{°C · kg}{mol}} = 2.5 \frac{mol}{kg}[/latex] = 2.5 m The molality of the solution is 2.5 m.

Question 11.11: USING BOILING-POINT ELEVATION TO CALCULATE THE MOLALITY OF A...

Chemistry

USING BOILING-POINT ELEVATION TO CALCULATE THE MOLALITY OF A SOLUTION

What is the molality of an aqueous glucose solution if the boiling point of the solution at 1 atm pressure is 101.27 °C? The molal boiling-point-elevation constant for water is given in Table 11.4.

TABLE 11.4 Molal Boiling-Point-Elevation Constants ( $K_{b}$ ) and Molal Freezing-Point-Depression Constants ( $K_{f}$ ) for Some Common Substances

Substance

K_{b}

[(°C · kg)/mol]

K_{f}

[(°C · kg)/mol]

Benzene (

C_{6}H_{6}

) 2.64 5.07
Camphor (

C_{10}H_{16}O

) 5.95 37.8
Chloroform (

CHCl_{3}

) 3.63 4.70
Diethyl ether (

C_{4}H_{10}O

) 2.02 1.79
Ethyl alcohol (

C_{2}H_{6}O

) 1.22 1.99
Water (

H_{2}O

) 0.51 1.86

STRATEGY
Rearrange the equation for molal boiling-point elevation to solve for m:

ΔT_{b}  =  K_{b}  ·  m               so                m  =  \frac{ΔT_{b}}{K_{b}}

where $K_{b}$ = 0.51 (°C · kg)/mol and $ΔT_{b}$ = 101.27 °C – 100.00 °C = 1.27 °C.

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