A solution, in which the magnesium concentratio is 50mg/L, is needed. The source

Question

A solution, in which the magnesium concentratio is 50mg/L, is needed. The source of magnesium ions is magnesium chloride. If 1.5L of the solution is needed, what weight of magnesium chloride must be weighed out? (FW of Mg is 24.305g/mol , and MgCl2 is 95.211g/mol.)

Part B:

if 10.00mL of the above solution (50mg/L mg2+) is pipeted into a 100mL volumetric flask and diluted to the mark with DI water, and then 5ml of that solution is pipeted into a 250 ml volumetrick flask and diluted to the mark with DI water, what is the concentration of magnesium in the final solution in ppb?

Explanation / Answer

A solution, in which the magnesium concentratio is 50mg/L, is needed. The source of magnesium ions is magnesium chloride. If 1.5L of the solution is needed, what weight of magnesium chloride must be weighed out? (FW of Mg is 24.305g/mol , and MgCl2 is 95.211g/mol.)

mass of Mg needed=50mg/L*1.5L=75mg

molar mass of MgCl2=95.21g/mole(containing Cl2=71g/mole)

1mole of MgCl2 contain=24.21g Mg2+=24210mgMg2+

24210mgMg2+ is present in=95210mgMgCl2

75mgMg2+ is present in=75*95210/24210=294.95MgCl2

Part B:

if 10.00mL of the above solution (50mg/L mg2+) is pipeted into a 100mL volumetric flask and diluted to the mark with DI water, and then 5ml of that solution is pipeted into a 250 ml volumetrick flask and diluted to the mark with DI water, what is the concentration of magnesium in the final solution in ppb?

Concentration of first diluted solution=

M1=M*V/V1=50mg/L*10mL/100mL=5mg/L

concentration of 2nd diluted solution

M2=M1V1/V2=5mg/L*5mL/250mL=0.1mg/L=(0.1*1000)microgram/L=100ppb

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