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Pepsi [2]
4 years ago
13

Urea is a common fertilizer with the formula CO(NH2)2 that is sometimes used as a chemical de-icer for icy road surfaces in the

winter months. Adding a solute to a solvent causes the freezing point to be lowered and allows the solvent to remain liquid at lower temperatures. The value of Kf for water is 1.86°C/m, and its normal freezing point is 0.00°C. Urea is a nonvolatile nonelectrolyte solute. What molality of urea is required to lower the freezing point of ice water by 1.13°C?
Chemistry
1 answer:
Lunna [17]4 years ago
4 0

Answer:

The answer to your question is  molality = 0.61

Explanation:

Freezing point is the temperature at which a liquid turns into a solid if a solute is added to a solution, the freezing point changes.

Data

Kf = 1.86 °C/m

molality = ?

ΔTc = 1.13°C

Formula

ΔTc = kcm

Solve for m

m = ΔTc/kc

Substitution

m = 1.13 / 1.86

Simplification and result

m = 0.61

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4 0
3 years ago
A compound is found to contain 7.523% phosphorus and 92.48% iodine by weight. what is the empirical formula for this compound?
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The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
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Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

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The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

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