Chemistry

# What is the molarity of H2C2O4 2H2O solution?

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###### 2014-01-20 19:04:02

molarity of known H2C2O4

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Molarity (concentration ) = moles of solute/Liters of solution 250.0 ml = 0.250 liters 2.431 grams H2C2O4 * 2H2O ( 1mole cpd/ 126.068 grams) = 0.01928 moles H2C2O4 * 2H2O Molarity = 0.01928 moles cpd/0.250 liters = 0.07712 Molarity

It is not important whether this '200 ml of solution' is countable without or without the hydrated compounds contribution of H2C2O4.2H2OAt first find moles of this3.78 grams H2C2O4 . 2H2O ( 1 mole cpd./126.068 grams)= 0.02998 moles H2C2O4 . 2H2OSince Molarity = moles of solute/Liters of solution ( 200 ml = 0.200 Liters )Molarity = 0.02998 moles H2C2O4 . 2H2O/0.200 Liters= 0.150 M H2C2O4.2H2O

0.063 g of oxalic acid * (1 mol H2C204*2H2O / 126.07 g) = 0.0004997 mol H2C2O4*2H2O 0.0004997 mol H2C204 / 0.250 L = 0.001999 M of H2C2O4

Molar mass H2C2O4.2H2O =126.0658 g/mol 2.38g = 2.38/126.0658 = 0.0189mol dissolved in 200ml solution. Molarity = 1000/200*0.0189 = 0.094M

molar mass H2C2O4x2H2O = 2+24+64+36 = 126 g/molemoles = 2.38 g x 1mol/126 g = 0.0189 molesmolarity = 0.0189 moles/0.200 L = 0.0944 M

Molarity solution is not a term that I have ever heard used. Molarity is always a term used to describe a concentration of solute in a solvent, i.e. of a solution. The closest I've heard to molarity solution is when people ask "what is the solution's molarity".

Molarity = moles of solute/Liters of solution Molarity = 5 moles solute/4.5 Liters of solution = 1 M solution ==========

Molarity of a solution depends upon volume of solution, the change in temperature changes the volume so molarity changes.

why molarity is preferred over molarity in expressing the concentration of a solution

Molarity= (number of moles of solute)/(volume of solution in dm3)

Molarity is an indication for concentration.

If you concentrate a solution, the molarity (moles/liter) will increase.

Sugar does not have measurable molarity. Molarity is used to determine the concentration of a solute in a solution. For example, you could measure the molarity of sugar in a sugar-water solution.

Adding more solvent to a solution decreases the molarity of the solution. This is based on the principle that initial volume times initial molarity must be equivalent to final volume times final molarity.

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