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3 years ago

1.2 g of sucrose is added to water with the final volume of 75 ml. What is the concentration of sucrose in %w/v?

Chemistry

1 answer:

Firlakuza [10]3 years ago

8 0

Answer:

16 g/L

Explanation:

Given that:-

Mass of sucrose added = 1.2 g

Volume of water = 75 mL = 0.075 L ( 1mL = 0.001 L )

%w/v is defined as the mass of the solution in 1 L of the solution.

So,

$\%\ w/v=\frac{Mass_{solute}}{Volume_{Solution}}$

$\%\ w/v=\frac{1.2\ g}{0.075\ L}=16\ g/L$

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a_sh-v [17]

Answer:

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Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

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