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4 years ago

Calculate the molality of a solution formed by adding 6.30 g NH4CL to 15.7 g of water

Chemistry

1 answer:

babymother [125]4 years ago

6 0

Answer:

Molality = 7.5 mol/kg

Explanation:

Given data:

Mass of NH₄Cl = 6.30 g

Mass of water = 15.7 g (15.7/1000 =0.016 kg)

Molality = ?

Solution:

Formula of molality:

Molality = Moles of solute / mass of solvent in gram

Now we will first calculate the number of moles of solute( NH₄Cl )

Number of moles = mass/ molar mass

Molar mass of NH₄Cl = 53.491 g/mol

Number of moles = 6.30 g/ 53.491 g/mol

Number of moles = 0.12 mol

Now we will calculate the molality.

Molality = Moles of solute / mass of solvent in gram

Molality = 0.12 mol / 0.016 kg

Molality = 7.5 m

or (m=mol/kg)

Molality = 7.5 mol/kg

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Thepotemich [5.8K]

Answer: The final temperature of the mixture is 49°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

For cold water:

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g$

For hot water:

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g$

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 120 g

$m_2$ = mass of cold water = 230 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of hot water = 95°C

$T_2$ = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]$

$T_{final}=49^oC$

Hence, the final temperature of the mixture is 49°C

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