How many grams of \(\mathrm{NaOH}\) are needed to prepare \(500.0 \mathrm{~mL}\) of \(0.300 \mathrm{M} \mathrm{NaOH}\) solution?

Short Answer

Expert verified
To prepare 500.0 mL of 0.300 M NaOH solution, 6.0 g of NaOH are needed.

Step by step solution

01

Understand the Problem and Gather Information

The problem is essentially asking to find out how many grams of NaOH are needed to prepare a 0.300 M solution of NaOH in a volume of 500.0 mL. The molar mass of NaOH is calculated by adding the atomic masses of sodium (Na - 23 g/mol), oxygen (O - 16 g/mol), and hydrogen (H - 1 g/mol), which results in 40 g/mol.
02

Convert Volume from mL to L

Since molarity is defined in terms of liters of solution, the given volume (500 ml) has to be converted to liters. This is done using the conversion factor of 1 L = 1000 ml, resulting in: \[ 500.0 mL * \frac{1 L}{1000 mL} = 0.5 L \]
03

Use the Molarity Formula to Find Moles of NaOH

Using the formula of molarity from earlier: \[Molarity = \frac{Moles~of~Solute}{Liters~of~Solution}\] the moles of NaOH can be found by rearranging the formula as follows: \[Moles~of~Solute = Molarity * Liters~of~Solution\] Then substituting the given molarity (0.300 M) and the converted volume (0.5 L): \[Moles~of~NaOH = 0.300 M * 0.5 L = 0.15 mol\]
04

Convert Moles to Grams using Molar Mass

The final step involves converting the moles of NaOH to grams. This is done by multiplying the moles of NaOH by the molar mass of NaOH (40 g/mol): \[Grams~of~NaOH = Moles~of~NaOH * Molar~Mass~of~NaOH = 0.15 mol * 40 g/mol = 6.0 g\] In conclusion, 6.0 g of NaOH are needed to prepare 500.0 mL of a 0.300 M NaOH solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free