Answer:
the osmolarity of the 0.85% NaCl solution is approximately equal to 300 mosmol/L.
Explanation:
Molarity= number of moles/volume of the solution in liters
volume of the solution= 100 mL
1 L= 1000 mL
100 mL= 100/1000=0.1 L
molar mass of NaCl= 58.44 g/mol
number of moles= mass in gram/gram molecular mass
= 0.85/58.44
= 0.01455
molarity= number of moles/volume of the solution in liters
= 0.01455 / 0.1
= 0.1455 M
1 M= 1000 mM
so 0.1455 M = 0.1455 × 1000
= 145.5 mM
Osmolarity is the concentration of solutes in the solution.
NaCl dissociates to Na+ and Cl-
so osmolarity of 145.5 mM NaCl= 2 × 145.5
= 291 mosmol/L
Therefore, we conclude that the osmolarity of the 0.85% NaCl solution is approximately equal to 300 mosmol/L.
Answer is: V has the units of liters, and T has the units of kelvin.
Ideal gas law: p·V = n·R·T.
atm · L = mol · L·atm/mol·K · K; both side of equatation have same values.
R = 0,08206 L·atm/mol·K; universal gas constant.
p is pressure of the gas, unit is standard atmosphere (atm).
V is volume of the gas, unit is liters (L).
n is amount of substance of the gas; unit is mole (mol).
T is temperature of the gas, unit is Kelvin (K).
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