7. In liquid diffusion experiment, you need to prepare NaCl solution by dissolvi

Question

7. In liquid diffusion experiment, you need to prepare NaCl solution by dissolving 5.85 mole/ m and the g of solid NaCl in 100 ml of water. Determine the concentration in mole fraction of NaCl in the solution. The molecular weight of NaCl is 58.4 g/mole and the molecular weight of water 18 g/mole. BI is nequined to caloulate the heat evolved from hot water fluid, from the mass flow rate and specific heat and temperature difference. F-18 Limin Given: density at 47'C-0.9982 g/am'; density at 20 °C-998.2 Kg/m p at 47 "C-4179.6JKg KOp at 20 "C-4.1816 Kj/kg.K 30%)

Explanation / Answer

Solution for question number 7:

1. As we know the molarity formula

Molarity = (Weight of the substance dissolved / M.Wt.)* 1000/ Vol.of the water in ml

So after substituting the values we will have

Molarity of NaCl = (5.85/ 58.4)*1000/100

                         = 1.0017

i.e. Molarity of NaCl solution is 1.0017Moles per Litre Or m3( 1 Litre is also called 1 metre cube)

2. Mole fraction formula for our given NaCl is

Mole fraction of NaCl = No. Of moles of NaCl/ Total number of moles(total no.of NaClmoles+total no.of water moles)

Hence from our given data

No.Of moles of NaCl = Weight of NaCl/M.Wt of NaCl

i.e.                          = 5.85/58.4

                              = 0.1 moles.

similarly for water

No. Of moles of water = Weight of the water ( 100 ml water means 100 grams of water because density of water is one) / M.Wt. of water

                              = 100/18

                              = 5.5555

Finally

Mole fraction of NaCl = Total moles of Nacl/ Sum of the moles of NaCl and Water

So

Mole fraction of NaCl is = 0.1/ (0.1+5.5555)

                                   = 0.0176819

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