Question

Only answer if you are 100% positive! Easy question, I think I'm just overthinking it. Make sure to include units, show work!

A. Saline is usually 0.85% (wt/vol) NaCl in di-H2O. The molecular weight of NaCl is 58. Whati s the molarity of saline?

B. A total dilution of 1000 was made using two separate steps. In the first step, 9.6 mL was placed into 48 mL. What volume from the first dilution was placed into 63 mL?

Answer 1

A.

MW of NaCl = 58 (given)

0.85% (wt/vol) of NaCl in water means 0.85 g of NaCl is dissolved in 100 mL of water

Number of moles of NaCl in 0.85 g = Weight/Mw = 0.85/58 = 0.014655 mole

Molarity = number of moles/Volume in Lit. = 0.014655 mole/0.100 L = 0.14655 mole/L

B.

First dilution:

M1 = M, V1 = 9.6 mL; M2 = ?, V2 = 9.6 mL + 48 mL = 57.6 mL

M1V1 = M2V2

M2 = M1V1/V2 = (M x 9.6)/57.6 = 0.167 M

Second dilution:

M2 = 0.167 M, V2 = ?; M3 = M/1000 =.001M, V3 = (V2 + 63) mL

M2V2 = M3V3

0.167 x V2 = 0.001 x (V2 + 63)

0.167 V2 = 0.001 V2 + 0.063

0.166 V2 = 0.063

V2 = 0.38 mL