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Calculate the amount of ice that will se...

Calculate the amount of ice that will separated out when a solution containing 50 g of ethylene glycol in 200 g of water is colled to `-9.3^(@)C." "(K_(f)" for "H_(2)O="1.86 K kg mol"^(-1))`

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To solve the problem of calculating the amount of ice that will separate out when a solution containing 50 g of ethylene glycol in 200 g of water is cooled to -9.3°C, we can follow these steps: ### Step 1: Calculate the moles of ethylene glycol (C2H6O2) - The molar mass of ethylene glycol (C2H6O2) is calculated as follows: - Carbon (C): 12.01 g/mol × 2 = 24.02 g/mol - Hydrogen (H): 1.01 g/mol × 6 = 6.06 g/mol - Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol - Total = 24.02 + 6.06 + 32.00 = 62.08 g/mol ...
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Calculate the amount of ice that will separate out on cooling containing 50 g of ethylene glycol in 200 g of water to -9.3^(@)C (K_(f) for water = 1.86 K mol^(-1) kg )

Calculate the amount of ice that will separate out on cooling a solution containing 50g of ethylene glycol in 200g water to -9.3^(@)C ( K_(f) for water = 1.86 "K kg mol"^(-1) )

The amount of ice that will separate out from a solution containing 25 g of ethylene glycol in 100 g of water that is cooled to -10^(@)C , will be [Given : K_(f) "for" H_(2)O = 1.86K mol^(-1) kg ]

The amount of ice that will separate out on cooling a solution containing 50 g ethylene glycol in 200 g water to -9.3^(@)C is : ( K'_(f) = 1.86 K "molality"^(-1) )

What will be the amount of ice separated on cooling solution of 40g ethylene glycol in 400 g water upto -9.3^(@) C ?

A solution containing 62 g ethylene glycol in 250 g water is cooled to -10^(@)C . If K_(f) for water is 1.86 K mol^(-1) , the amount of water (in g) separated as ice is :

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