75 mL of PBS buffer needed All 6 components are solid and the grams need to be d

Question

75 mL of PBS buffer needed

All 6 components are solid and the grams need to be determined with the exception fo the CaCl2 that is a 10x liquid solution(determine mL that will be needed). mw =molecular weight

137mM NaCl mw 58.44g/mol

2.7mM KCL mw 74.5513 g/mol

8.1mM Na2HPO4 mw 141.96 g/mol

1.8 KH2 PO4 mw 136.086 g/mol

.01% CaCl2 g/mol mw 110.98 g/mol

.01% MgCl2*6H20 mw 203.20 g/mol

Am i doing the stoichemtry correctly as I get a really small number and do not have a scale that will messure this low?

1) NaCL

(.000137mol/L)(.075L/1)(58.44g/mol)= 6.0047x10^-4 grams NaCL needed

For the liquid I am confused on how to calculate the 10x liquid solution.

Please work this problem out for me along with an explanation if I did my calculations incorrect above.

(.01g/.1L)(mol/110.98g)=9.01x10^-5 M x 10 =? Confused

Thank You

Explanation / Answer

grams for NaCl are calculated correctly

0.01 % CaCl2 means 0.01 g of CaCl2 solute are dissolved in in 100 g of solution. It implies, in 1000 ml [1L] of solution, there ate 0.1 g of CaCl2

10x liquid solution means, 10 times the given concnetartion

so the 10x solution has (0.1 g of CaCl2 in 1 L) * 10 times

= 1g of CaCl2 in 1 Litre solution

Molarity = grams/molar mass

Volume in litres

Molarity of CaCl2 of 10 soltuion = 1g / (110.98 g/mole)

1 litre

= 0.009 moles / 1 litre

so concentration of 10x CaCl2 solution = 0.009 M

0.01 % MgCl2*6H2o means 0.01 g of salt dissolved to make 100 ml of salt solution

==> 0.1 g in 1000 ml [1 L] solution

so 0.1g in 1 L solution

so Molarity = 0.1g / 203.20

1 litre

= 4.921 * 10-4 M

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