Questions and Applications: You must show complete calculations, including set u

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Questions and Applications: You must show complete calculations, including set up, units, an correct significant figures to receive credit for problems requiring a numeral response. requiring an explanation must be written in complete sentences using no personal pronouns, contractions or abbreviations Answers Sodium chloride can be recovered from salt-water by evaporating the water. The Great Salt Lake in Utah contains salt-water which averages about 17.5% m/m NaCl. If 1.0L of salt- water from the Great Salt Lake is heated and all the water is removed, 205g of salt is recovered. What was the mass of water removed in the heating process? 1. nsInc! 263 2. If the concentration of NaCl in a solution is 2.5M what is the concentration in units of mass/volume%? I L A student is asked to prepare 750mL of an aqueous 0.18M NaOH solution. The student is provided with water, solid NaOH, a scale, a graduated cylinder and a container which holds exactly 750mL. Explain the steps required (including calculations!) to prepare this solution. 3. a. If the student was asked to prepare this same solution (750mL of 0.18M NaOH) using a concentrated 6.0M NaOH solution instead of solid NaOH and water, explain the steps (including calculations) the student would take to make the solution.

Explanation / Answer

Solution:1).

We have given In salt water NaCl M/M % is 17.5

Concept:M/M % is ,mass of solute present in 100 gram solution

So 17.5 M/M % means 17.5 gram NaCl present in 100 gram solution. Means water is 100-17.5=82.5 gram.

Now since 17.5 gram NaCl present in 82.5 gram water.

So 1 gram Nacl present in 82.5/17.5=4.71 gram water

so 205 gram NaCl will present in (82.5/17.5)×205=966.43 gram

So the mass of water removed in heating process is 966.43 gram.

Solution 2)

Given Concentration of NaCl=2.5M

We Know that Concentration=(Number of mole of solute/Volume of Solution in Litre)

Molecular Weight of NaCl=23+35.5=58.5 And Number of mole of NaCl=Mass of NaCl/58.5

volume should be l=v/1000

So Number of mole of NaCl/volume=(Mass of NaCl)/(58.5×V)=2.5

Mass/Volume=(2.5×58.5)/1000=0.14625

So (Mass/Volume)%=14.625%

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