Question

4. The boiling point elevation of 0.6 g acetic acid in 100 g benzene is 0.1265K. What conclusion

can you draw about the molecular state of the solute in solution? Molal boiling point constant

for benzene is 2.53 degree per mole​

Answer 1

Answer:

Here

W1=100g

W2=0.30g

ΔTb=0.0633K

Kb=2.53Kkgmol−1

∴ Molar mass of acetic acid M2=Kb×W2×1000/ΔTb×W1

⇒2.53×0.30×1000/0.0633×100

⇒119.90

Van't Hoff factor

i=Normal molar mass/Abnormal molar mass

=60/119.90

=0.5

Hence i<1 ,therefore the solute (acetic acid) is associated.

Explanation:

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here

W1=100g

W2=0.30g

टीबी =0.0633K

केबी =2.53Kkgmol

एसिटिक एसिड एम 2= केबी ×W2×1000/एडीटर टीबी ×1 का एक मोलर द्रव्यमान।

⇒2.53×0.30×1000/0.0633×100।

⇒119.90

नहीं हफ़ कारक कारक

= सामान्य दाढ़ द्रव्यमान/असामान्य दाढ़

=60/119.90

=0.5

इसलिए मैं <1, इसलिए सॉल्ट (एसिटिक एसिड) जुड़ा हुआ है।

Answer 2

Answer:

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Explanation:

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