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Nadya [2.5K]
3 years ago
8

Science: Number 3 plsss

Chemistry
2 answers:
Fynjy0 [20]3 years ago
7 0

Answer:

d

Explanation:

Tema [17]3 years ago
3 0

Answer:

the plants it will be bigger if you take care it so much the plant it will help in our world

Explanation:

I HOPE ITS HELP<em>.</em><em>.</em><em>.</em><em>P</em><em>L</em><em>E</em><em>A</em><em>S</em><em>E</em><em> </em><em>C</em><em>O</em><em>R</em><em>R</em><em>E</em><em>C</em><em>T</em><em> </em><em>M</em><em>E</em><em> </em><em>I</em><em>F</em><em> </em><em>I</em><em>M</em><em> </em><em>W</em><em>R</em><em>O</em><em>N</em><em>G</em>

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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
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