Question

1. Calculate the percent strength of a 200mL solution containing 350 milliosmoles of NaCl. (MW of NaCl = 58.5)?
2. You are trying to prepare 730 mL of a solution containing 145 mEq/L of NaCl (MW = 58.5), how many milligrams of NaCl should be used in preparing that solution?
3. Calculate the number of milliosmoles of anhydrous dextrose in 2750 mL of 7% w/v solution of anhydrous dextrose (MW = 180) in water?

Answer 1

-We have,

Percentage strength=( mass of solute/volume)*100

NaClis 350 millimoles, that is 0.35 moles.

percentage strength = (0.35*58.5/200)*100

= 10.23%

- Given that, there is 145 mEq in 1 litre (1000ml)

So, there will be 105.85 mEq in 730 ml

mg of NaCl= (molecular weight*mEq)/valancy of Nacl

=(58.5*105.85)/2

=3096mg or 3.096g of nacl

-7%W/V of dextrose means 7grams in 100ml

so in 2750 ml, there will be (7/100)*2750=192.5 grams of dextrose

so, number of millimoles of anhydrous dextrose = (192.5*180)/1000= 34.65 millimoles