Question

شما
18 g of glucose, C,H,O, is dissolved in 1 kg of water in a saucepan. Exa
At what temperature will water boil at 1.013 bar? K, for water is 0.52
K kg mol
Moles of clucose = 18 g/ 180 g mol = 0.1 mol
Solo​

Answer 1

Answer:

Heya user........❤

Using this formula, we get

$delta \: T \: = Kb \: \times (wB \times 1000) \div (mB\times wA)$

=> delta T = 0.52 ×( 18 × 1000) / (180× 1000)

=> T - T0 = 0.052

=> T = 373. 052 K

Answer 2

Answer:

y have u any other account.........

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