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Andreas93 [3]
3 years ago
5

What is the molality of a solution that has 3 mol of glucose in 6 kg of water?

Chemistry
2 answers:
Cloud [144]3 years ago
4 0
<span>it is 3/6 = 0.5 mol/kg of solvent i.e. 0.5 molal <span>solution</span></span>
Rudik [331]3 years ago
3 0

Answer: 0.5 m

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

Molality=\frac{n\times 1000}{W_s}

where,

n= moles of solute

Moles of glucose = 3 moles

W_s = weight of solvent (water) in g  = 6 kg = 6000 g   (1kg=1000g)

Now put all the given values in the formula of molality, we get

Molality=\frac{3moles\times 1000}{6000g}=0.5mole/kg

Therefore, the molality of solution will be 0.5 mole/kg.

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A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
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Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

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

Mass of the piece of metal = 80.0 grams

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Initial temperature of the metal = 88.0 °C

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Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

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with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

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