Question

When a 7.10 g sample of lithium nitrate dissolves in 28.00 g of water in a coffee-cup calorimeter, the temperature rose from 25.20 °C to 27.22 °C. Assume that the solution has the same specific heat as liquid water, i.e., 4.184 J/g.°C. Give the equation for the dissolution of solid lithium nitrate in water.

Answer 1

Answer on Question #60384 - Chemistry - General Chemistry

Question:

When a 7.10 g sample of lithium nitrate dissolves in 28.00 g of water in a coffee-cup calorimeter, the temperature rose from 25.20 °C to 27.22 °C. Assume that the solution has the same specific heat as liquid water, i.e., 4.184 J/g.°C. Give the equation for the dissolution of solid lithium nitrate in water.

Solution:

1) Write down the generic equation:

$$\mathrm{LiNO_3} + \mathrm{H_2O} = \mathrm{Li^+_{aq}} + \mathrm{NO_{3^-_{aq}}} + \mathrm{H_2O} + \mathrm{Q},$$

Where Q – is heat of dissolution.

2) We have to find the amount of heat per 1 mole of compound.

The total amount of heat in experiment:

$Q_{\text{total}} = \Delta T$ (temperature change)\*m solution (mass of solution)\*c (specific heat).

$Q_{\text{total}} = (27.22^{\circ}\mathrm{C} - 25.20^{\circ}\mathrm{C}) \times (7.10\mathrm{g} + 28.00\mathrm{g}) \times 4.184\mathrm{J/g}^{\circ}\mathrm{C} = 296.65\mathrm{J}.$

The heat of dissolution (per mole)

$Q = (Q_{\text{total}} / m$ (mass of compound))\*M (molar mass of compound).

$$\mathrm{M_{LiNO_3}} = 6.94 + 14.01 + 3 \times (16.00) = 68.95 \mathrm{~g/mol}$$ $$Q = (296.65 \mathrm{~J} / 7.10 \mathrm{~g}) \times 68.95 \mathrm{~g/mol} = 2880.85 \mathrm{~J/mol} (2.88 \mathrm{~kJ/mol})$$

Answer:

$$\mathrm{LiNO_3} + \mathrm{H_2O} = \mathrm{Li^+_{aq}} + \mathrm{NO_{3^-_{aq}}} + \mathrm{H_2O} + 2.88 \mathrm{~kJ}$$

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