Question #168141


What mass of NaCl is needed to produce a 500 g of a 6.40% M/M solution?


What is the concentration, in g/100 mL, of a solution made by dissolving 123.2 g of phosphoric acid in 50.0 mL of solution?


How much water is needed to produce a 0.68 M solution of copper (I) carbonate, if 3.58 g is used?


What mass of carbon dioxide would you add to 1 500 L of water in order to make a 325 ppm solution?


What is the concentration of a solution made by diluting 14.0 mL of a 4.20 M Na2CO3 to 86.0 mL?


To what volume should 58.3 mL of a 1.4 M KCl solution be diluted to produce a 0.45 M solution?


What volume of a 6.0 M NaOH solution is needed to produce 3.0 L of a 0.1 M solution?


What mass of aluminum nitrate do you need to prepare 3.58 L of a 1.77 M solution?


Zinc iodide is dissolved in 2.5 L of water to make a solution. 1 L of this solution is then diluted to make 12.0 L of a 0.5 M solution. What mass of zinc iodide was added to make the original solution?


Explain the steps you would follow to make 500 mL of a 0.3 M solution of sucrose (C12H22O11) in the lab


1
Expert's answer
2021-03-03T01:55:08-0500

1. %m/m of asolution=solute masssolution mass×100\%m/m \ of \ a solution=\dfrac{solute \ mass }{solution\ mass}×100

6.4=solute mass500×1006.4=\dfrac{solute\ mass}{500}×100

Mass of NaCl required=32gNaCl\ required=32g

2. Concentration of a solution=123.2g50ml=x100=\dfrac{ 123.2g}{50ml}=\dfrac{x}{100}

=123.2100=12320g/ml=123.2*100=12320g/ml

x=242.6gx=242.6g

3.

Molarity=molesvolume(L)=\dfrac{moles}{ volume(L)}

Moles of copper (i) carbonate =massmolarmass=\dfrac{mass}{ molar mass}


=3.58187=0.019=\dfrac{3.58}{187}=0.019


0.68=0.019L0.68=\dfrac{0.019}{L}

L=281.54mLL=281.54mL

4.

1ppm=1mg/L in aq solutionsFor325ppm=325mg/LTherefore,1500L325mg=487/500mg Mass=488g of CO2 required1ppm=1mg/L\ in\ aq\ solutions\newline For 325ppm=325mg/L\newline Therefore,\newline 1500L*325mg=487/500mg \ Mass=488g\ of\ CO_2 \ required

5.

C1V1=C2V2C186=4.2614C1=0.68=0.68MC1V1=C2V2\newline C1*86=4.26*14 \newline C1=0.68\newline =0.68M

6.

58.3ml0.4M0.45M=181mL\dfrac{58.3ml*0.4M}{0.45M}=181mL

7.

M1V1=M2V26.0V1=0.13.0V1=0.13.06.0M1V1=M2V2\newline 6.0*V1=0.1*3.0\newline V1=\dfrac{0.1*3.0}{6.0}


V1=0.05L of NaOH is neededV1=0.05L \ of\ NaOH\ is\ needed






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