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zzz [600]
3 years ago
10

What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m

Chemistry
1 answer:
bulgar [2K]3 years ago
4 0

Answer:

T_f for given question is 2.79 and T_b is 0.52

\Delta T_b = I \times K_b \times m {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

\Delta T_b = I \times K_b \times (w/M)/ x

\Delta T_b = I \times K_b \times w/Mx

\Delta T_b = 1 \times 0.51 \times1.5/(0.250 \times 58.44) = 0.052

\Delta T_f = M \times K_f = 1.86 \times 1.5 = 2.79

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<u>Step 1:</u> Data given

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<u>Step 2:</u> The balanced equation:

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<u>Step 3:</u> Calculate moles of Al

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Moles Al = 0.0678 moles

<u>Step 4:</u> Calculate moles of NaOH

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Moles NaOH = 0.1075 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

<u>Step 6</u>: Calculate moles of hydrogen

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Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

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