At 300 K, 36 g of glucose present per litre in its solution has an osm

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pi=CRT" (C = molar concentration)" (pi(1))/(pi(2))=(C(1))/(C(2))," "(4.98)/(1.52)=(36//180)/(C(2))" or "C(2)=(36)/(180)xx(1.52)/(4.98)="0.061 M"

please explain the question and tell me what is 4 98 bar in this question and why it's not been - Chemistry - Solutions - 14451181

At `300 K`, `36 g` of glucose present per litre in its solution has an osmotic pressure of `4.98

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At `300K,36g` of glucose present per litre in its solution had an osmotic pressure `4.98 ` bar. If